Problem: Simplify; express your answer in exponential form. Assume $n\neq 0, p\neq 0$. $\dfrac{{(n^{-4}p^{3})^{-1}}}{{(np^{-3})^{-3}}}$
To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(n^{-4}p^{3})^{-1} = (n^{-4})^{-1}(p^{3})^{-1}}$ On the left, we have ${n^{-4}}$ to the exponent ${-1}$ . Now ${-4 \times -1 = 4}$ , so ${(n^{-4})^{-1} = n^{4}}$ Apply the ideas above to simplify the equation. $\dfrac{{(n^{-4}p^{3})^{-1}}}{{(np^{-3})^{-3}}} = \dfrac{{n^{4}p^{-3}}}{{n^{-3}p^{9}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{4}p^{-3}}}{{n^{-3}p^{9}}} = \dfrac{{n^{4}}}{{n^{-3}}} \cdot \dfrac{{p^{-3}}}{{p^{9}}} = n^{{4} - {(-3)}} \cdot p^{{-3} - {9}} = n^{7}p^{-12}$